4th order Butterworth HPF lossless ladder

Approach. Take the normalized 4th order Butterworth LPF ladder ($R_1 = R_2 = 1\ \Omega$, $\omega_c = 1$ rad/s) from the table and apply the LP to HP transformation $s \to 1/s$.

Step 1: LP prototype (from table, $n = 4$)

Element values: $g_1 = 0.7654$, $g_2 = 1.848$, $g_3 = 1.848$, $g_4 = 0.7654$. Using the series-L first (T) form:

$L_1 = 0.7654$ H (series), $C_2 = 1.848$ F (shunt), $L_3 = 1.848$ H (series), $C_4 = 0.7654$ F (shunt).

These come from $t(s) = 1/B_4(s)$ with $B_4(s) = (s^2+0.7654s+1)(s^2+1.848s+1)$, $\rho = s^4/B_4(s)$ and the continued fraction expansion of $Z_1 = (1+\rho)/(1-\rho)$.

Step 2: LP to HP transformation

Replacing $s$ by $1/s$:

• series $L$ (impedance $sL$) becomes $L/s$, a series capacitor $C = 1/L$;
• shunt $C$ (admittance $sC$) becomes $C/s$, a shunt inductor $L = 1/C$;
• terminations are unchanged.

Step 3: Circuit

 R1=1  C1=1.3066 F  C3=0.5411 F
 o-/\/\---||----+----||---+------+
 |              |         |      |
(~)Vs          L2        L4     R2=1
 |              |         |      |
 o--------------+---------+------+
 L2 = 0.5411 H, L4 = 1.3066 H

The response is $\left|\dfrac{V_2}{V_s}\right| = \dfrac{0.5}{\sqrt{1+(1/\omega)^8}}$: 0 at dc, $-3$ dB (relative) at 1 rad/s and 0.5 at high frequencies.

Answer (normalized, $\omega_c = 1$ rad/s): $R_1 = 1\ \Omega$, series $C_1 = 1.3066$ F, shunt $L_2 = 0.5411$ H, series $C_3 = 0.5411$ F, shunt $L_4 = 1.3066$ H, $R_2 = 1\ \Omega$. For a cutoff $\omega_c$ and termination $R$: $C' = C/(R\omega_c)$, $L' = LR/\omega_c$. (Starting from the dual $\pi$ form gives shunt $L_1 = 1.3066$ H, series $C_2 = 0.5411$ F, shunt $L_3 = 0.5411$ H, series $C_4 = 1.3066$ F.)